Critical angle for TIR in a glass prism with n=1.44

[Alain12345]

I completed a question about TIR in a glass prism, and I was wondering if someone can tell me if I got it right. Thanks

"Less" and "more" refer to the less dense medium and more dense medium, and the red line is the normal.

http://img145.imageshack.us/img145/7061/tirquestion0mi.png

Question: Determine all the necessary angles and then calculate at which angle the ray exits the prism. n= 1.44

Calculations:

n= sin less/sin more
1.44= sin 62/sin more
sin 62/1.44= sin more
angle more= 38
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n= sin less/sin more
1.44= sin less/ sin 60
angle less= N/A

therefore, sin C= 1/n
sin C= 1/1.44
= 0.6944
Critical angle= 44

angle more is greater than critical angle, therefore TIR occurs
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n= sin less/sin more
1.44= sin less/ sin 20
angle less= 30

The ray exits the prism at 30 degrees.

EDIT: I put the 62 in the wrong place, I meant it to be in between the normal and ray

 

[lightgrav]

The angle that is used in Snell's Law : sin(theta)
is the angle measured from the NORMAL line to the RAY.
The 62 degree angle, according to your diagram,
is measured from the SURFACE to the ray.

 

[Alain12345]

I'm sorry, I put the 62 in the wrong place when I was makign the diagram. It should be between the normal and the ray

 

[lightgrav]

Then your work is Okay.
You don't really need to compute the critical angle (paragraph 3)
... if the sin(less) is supposed to be greater than 1,
then that "less" angle does not exist, and Tot.Internal Reflection occurs.

 

[Alain12345]

So 30 degrees is okay for my final answer? Thanks a lot!